Question: What are the first three non-zero terms of the Maclaurin series for the function $f(x)=x\cos(x)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ x+x^2+\frac{x^3}{2!}$ (Choice B) B $ x-\frac{{{x}^{4}}}{3!}+\frac{{{x}^{6}}}{5!}$ (Choice C) C $ x-\frac{{{x}^{3}}}{2!}+\frac{{{x}^{5}}}{4!}$ (Choice D) D $ x+\frac{{{x}^{3}}}{2!}-\frac{{{x}^{5}}}{4!}$
Explanation: Start with the Maclaurin series for $\cos(x)$. $\cos x=1-\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{2n}}}{\left( 2n \right)!}+...$ Multiply the series by $x$. $ x\cos x=x-\frac{{{x}^{3}}}{2!}+\frac{{{x}^{5}}}{4!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{2n+1}}}{\left( 2n \right)!}+...$ The first three terms are $x-\dfrac{{{x}^{3}}}{2!}+\dfrac{{{x}^{5}}}{4!}$.